Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 7, Problem 73P
To determine

The direction and speed of the acrobats after they grab each other.

Expert Solution & Answer
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Answer to Problem 73P

The final velocity is 0.64m/s directed 73° with the horizontal.

Explanation of Solution

The speed of the acrobat with mass 60kg is 3.0m/s at an angle 10° above the horizontal. The speed of the acrobat with mass 80kg is 2.0m/s at an angle 20° above the horizontal.

Since finally the two acrobats grab each other, the collision is perfectly elastic and the final velocity of both the acrobats will be same.

Write the law of conservation of linear momentum of the acrobats in x-direction.

m1vi1x+m2vi2x=vfx(m1+m2)

Here, m1 is the mass of the first acrobat, vi1x is the initial velocity of m1 in x-direction, m2 is the mass of the second acrobat, vi2x is the initial velocity of m2 in x-direction, vfx is the final velocity in x-direction.

Re-write the above equation to get an expression for vfx.

vfx=m1vi1x+m2vi2xm1+m2 (I)

Write the law of conservation of linear momentum of the acrobats in y-direction.

m1vi1y+m2vi2y=vfy(m1+m2)

Here, vi1y is the initial velocity of m1 in y-direction, vi2y is the initial velocity of m2 in y-direction, vfy is the final velocity in y-direction.

Re-write the above equation to get an expression for vfy.

vfy=m1vi1y+m2vi2ym1+m2 (II)

Write the formula for the magnitude of the final velocity.

v=vfx2+vfy2 (III)

Here, v is the magnitude of the final velocity.

Write the formula for the direction of the final velocity.

θ=tan1(vfyvfx) (IV)

Here, θ is the angle made by the final velocity with the horizontal.

Conclusion:

Substitute 60kg for m1, 80kg for m2, (3.0m/s)(cos10°) for vi1x, (2.0m/s)(cos160°) for vi2x in equation (I).

vfx=(60kg)(3.0m/s)(cos10°)+(80kg)(2.0m/s)(cos160°)60kg+80kg=0.192m/s

Substitute 60kg for m1, 80kg for m2, (3.0m/s)(sin10°) for vi1y, (2.0m/s)(sin160°) for vi2y in equation (II).

vfy=(60kg)(3.0m/s)(sin10°)+(80kg)(2.0m/s)(sin160°)60kg+80kg=0.614m/s

Substitute 0.192m/s for vfx, 0.614m/s for vfy in equation (III).

v=(0.192m/s)2+(0.614m/s)2=0.64m/s

Substitute 0.192m/s for vfx, 0.614m/s for vfy in equation (IV).

θ=tan1(0.614m/s0.192m/s)=73°

The final velocity is 0.64m/s directed 73° with the horizontal.

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